1. The number of prime factors in the expressions 6^{4} × 8^{6} × 10^{8} × 12^{10} is:

a) 80

b) 64

c) 72

d) 48

Explanation: 6

^{4}× 8

^{6}× 10

^{8}× 12

^{10}

= (2 × 3)

^{4}× (2

^{3})

^{6}× (2 × 5)

^{8}× (2

^{2}× 3 )

^{10}

= 2

^{4}× 3

^{4}× 2

^{18}× 2

^{8}× 5

^{8}× 2

^{20}× 3

^{10}

= 2

^{50}× 3

^{14}× 5

^{8}

The total prime factors,

= 50 + 14 + 8 [By adding maximum power of prime factors.]

= 72

2. x is five digit number. The digit in ten thousands place is 1. the number formed by its digits in units and ten places is divisible by 4. The sum of all the digits is divisible by 3. If 5 and 7 also divide x, then x will be.

a) 14020

b) 12060

c) 10020

d) 10080

Explanation: Let the digits of x be

x = abcde

x = 1bcde [Given ten thousands place is 1.]

Now we can check the options as given that the sum of the all digit is divisible by 3.

10080 is the only number given in the option which satisfies all the given conditions.

3. A man sells chocolates which are in the boxes. Only either full box or half a box of chocolates can be purchased from him. A customer comes and buys half the number of boxes which the seller had plus half box more. A second customer comes and purchases half the remaining number of boxes plus half a box. After this the seller is left with no chocolate boxes. How many chocolate boxes the seller had initially?

a) 2

b) 3

c) 4

d) 3.5

Explanation: The best way to go through the options

Let there are initially 3 boxes then,

1

^{st}customer gets = $$\frac{3}{2}$$ + $$\frac{1}{2}$$ = 2

Remaining boxes = 3 - 2 = 1

2

^{nd}customer = $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 1

Option 'b' is correct.

4. If x + y + z = 0, then x^{3} + y^{3} + z^{3} is equal to :

a) 0

b) 3xyz

c) $$\frac{{{\text{xy}} + {\text{yz}} + {\text{zx}}}}{{{\text{xyz}}}}$$

d) xyz(xy + yz + zx)

Explanation:

x + y + z = 0

Cubing both side,

(x + y + z)^{3} = 0

x^{3} + y^{3} + z^{3} - 3xyz = 0 [using formula]

x^{3} + y^{3} + z^{3} = 3xyz

5. To write all the page numbers of a book, exactly 136 times digit 1 has been used. Find the number of pages in the book.

a) 190

b) 195

c) 210

d) 220

Explanation: From 1- 99 digit 1 is used 20 times. And From 100 - 199, 1 is used 120 times

So, from 1 to 199, 1 is used,

20 + 120 = 140 times

We need 136. So leave 199, 198, 197 and 196

Required pages = 195

6. Given, N = 98765432109876543210 ..... up to 1000 digits, find the smallest natural number n such that N + n is divisible by 11.

a) 2

b) 3

c) 4

d) 5

Explanation: For a no. to be divisible by 11,

Sum(odd digit nos) - Sum(even digit nos) = 0 or divisible by 11

If we look at 9876543210, the difference we get is 5

i.e. [(9 + 7 + 5 + 3 + 1) - (8 + 6 + 4 + 2 + 0) = 5]

The series is up to 1000 digit,

That means, $$\frac{{1000}}{{10}}$$ = 100 time 5,

then the difference will be 5 × 100 = 500

In order for the difference to be divisible by 11, we need to add 5 and the no will become 505

505 is divisible by 11

7. The remainder when 4^{0} + 4^{1} + 4^{2} + 4^{3} + ........ + 4^{40} is divided by 17 is:

a) 0

b) 16

c) 4

d) None of these

Explanation: Let S be the sum of the expression,

S = 4

^{0}+ 4

^{1}+ 4

^{2}+ 4

^{3}+ ........ + 4

^{40}

S = (1 + 4 + 16 + 64) + 4

^{4}(1 = 4 + 16 + 64) + ...... + 4

^{36}+ 4

^{40}

Since, (1 + 4 + 16 + 64) = 85, is divisible by 17. Hence, except 4

^{40}remaining expression is divisible by 17.

$$\frac{{{4^{40}}}}{{17}} \to \frac{{{{\left( {{4^4}} \right)}^{10}}}}{{17}} \to \frac{{{1^{10}}}}{{17}}$$

Remainder = 1

8. The remainder when 3^{0} + 3^{1} + 3^{2} + 3^{3} + . . . . . . . + 3^{200} is divided by 13 is:

a) 0

b) 4

c) 3

d) 12

Explanation:

$$\eqalign{ & {\text{The}}\,{\text{given}}\,{\text{expression}}\,{\text{is}}\,{\text{in}}\,{\text{GP}}\,{\text{series}} \cr & S = {3^0} + {3^1} + {3^2} + {3^3} + ........ + {3^{200}} \cr & S = {\frac{{ {{3^0} \times \left( {{3^{201}} - 1} \right)} }}{{ {3 - 1} }}} \cr & S = \frac{{ {{3^{201}} - 1} }}{2} \cr & S = \frac{{ {{{\left( {{3^3}} \right)}^{67}} - {1^3}} }}{2} \cr & S = \frac{{ {{{27}^{67}} - {1^3}} }}{2} \cr} $$

Since, (A

^{n}- B

^{n}) is divisible by (A - B), So, (27

^{67}- 1

^{3}) is divisible by (27 - 1) = 26

Hence, Expression is also divisible by 13 as it is divisible by 26

Given expression is divisible by 13 so the remainder will be 0

9. The distance between the house of Rajan and Raman is 900 km and the house of former is at 100^{th} milestone where as the house of Raman is at 1000^{th} milestone. There are total 901 milestone at a regular interval of 1 km each. When you go to Raman's house from the house of Rajan which are on same highway, you will find that if the last digit (i.e. unit digit) of 3 digit number on every milestone is same as the first (i.e. hundreds digit) of the number on the next milestone is same, then these milestones must be red colour and rest will be of black. Total number of red colour milestone is:

a) 179

b) 90

c) 189

d) 100

Explanation: Rajan (100)____________________ (1000) Raman

These numbers are:

(101, 102), (111, 102), (121, 122), (131,132).........(202, 203), (212, 213), ....... (303, 304), (313, 314) ......... (808, 809)......... (989, 990)

There are total 20 × 7 + 21 + 18 = 179, mile stones which are red, because the hundreds digit on the next milestone is same as the unit digit of previous milestone.

10. Half way through the journey from Delhi to Lahore Atalji began to look out of the window of the Samjhauta Express and continued it until the distance which was remained to cover was half of what he has covered. Now at this time how much distance he has to cover?

a) $$\frac{2}{5}$$

b) $$\frac{1}{4}$$

c) $$\frac{1}{3}$$

d) $$\frac{1}{6}$$

Explanation: Since, he has covered twice the distance which yet he has to cover. It means he has covered $$\frac{2}{3}$$ of the whole journey and remaining journey is $$\frac{1}{3}$$ rd.